POJ 3187-Backward Digit Sums【全排列+找规律】_halcon找规律个数物件-优快云博客

本文详细介绍了如何解决BackwardDigitSums问题，通过全排列和特定的数学规律求解给定序列的起始序列，同时提供了算法的C++实现。

Backward Digit Sums

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5669		Accepted: 3283

Description

FJ and his cows enjoy playing a mental game. They write down the numbers from 1 to N (1 <= N <= 10) in a certain order and then sum adjacent numbers to produce a new list with one fewer number. They repeat this until only a single number is left. For example, one instance of the game (when N=4) might go like this:

    3   1   2   4

      4   3   6

        7   9

         16

Behind FJ's back, the cows have started playing a more difficult game, in which they try to determine the starting sequence from only the final total and the number N. Unfortunately, the game is a bit above FJ's mental arithmetic capabilities.

Write a program to help FJ play the game and keep up with the cows.

Input

Line 1: Two space-separated integers: N and the final sum.

Output

Line 1: An ordering of the integers 1..N that leads to the given sum. If there are multiple solutions, choose the one that is lexicographically least, i.e., that puts smaller numbers first.

Sample Input

4 16

Sample Output

3 1 2 4

Hint

Explanation of the sample:

There are other possible sequences, such as 3 2 1 4, but 3 1 2 4 is the lexicographically smallest.

解题思路：

题目大意就是给出N，第一行是1到N这些数组成的一个序列，两两相加到最后有一个和，我们会给出这个和S，需要求的是第一行的这个序列，字典序最小。还是全排列，只不过我们需要找到其中的规律，

1 n=1

1 1 n=2

1 2 1n=3

1 3 3 1n=4

1 4 6 4 1

1 5 10 10 5 1

.。。。。。

由此看出我们只需要将每次全排列后的结果，乘以对应n中位置上的个数，就能得到整个三角形相加的和。

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int wc[100000];
int oo[100][100];
int main()
{
	int  n,m;
	int i,j;
	oo[1][0]=1;
	oo[2][0]=1;
	oo[2][1]=1;
	for(i=3;i<=10;i++)
	{
		oo[i][0]=oo[i][i-1]=1;
		for(j=1;j<i-1;j++)
		{
			oo[i][j]=oo[i-1][j-1]+oo[i-1][j];
		}
	}
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		
		for(i=0;i<n;i++)
		{
			wc[i]=i+1;
		}
		do{ 
		int ji=0;
			for(i=0;i<n;i++)
			{
				ji+=wc[i]*oo[n][i];
			}
            if(ji==m)  
            {  
               	for(i=0;i<n;i++)
               	{
               		if(i==n-1)
               		printf("%d\n",wc[i]);
               		else
               		printf("%d ",wc[i]);
               	}
                break;  
            }  
		}while(next_permutation(wc,wc+n));
	}
	return 0;
}